subgroups of finitely generated groups with a finite index Let $G$ be a finitely generated group and $H$ a subgroup of $G$. If the index of $H$ in $G$ is finite, show that $H$ is also finitely generated.
 A: Well, the standard argument is as follows. 
Let 
$$
g \mapsto [g] \qquad (g \in G)
$$
be a function which is constant on all right cosets of $H,$ and we require
$$
[e]=e.
$$
It is easy to see that
$$
u [u]^{-1} \in H, \quad [[u]]=[u], \quad [[u]v]=[uv] \qquad (u,v \in G)  \qquad \qquad (*)
$$
Now let
$$
S = \{ [g] : g \in G\}
$$
and $Y=Y^{-1}$ be a symmetric generating set of $G.$ Then the set
$$
\{ s y [sy]^{-1} : s \in S, y \in Y\}
$$
is a generating set of $H$ (a finite one, if both $S$ and $Y$ are finite $\iff$ the index of $H$ in $G$ is finite and $G$ is finitely generated).
For suppose that a product $y_1 \ldots y_r$ is in $H$ where $y_k \in Y$ ($k=1,\ldots,r$). Let, for example's sake, $r=3.$ Then
$$
y_1 y_2 y_3 = y_1 [y_1]^{-1} \cdot [y_1] y_2 [[y_1] y_2]^{-1} \cdot [[y_1] y_2] y_3
[[[y_1] y_2] y_3]^{-1} \qquad \qquad (**)
$$
where in the right hand side we have a product of elements of $H$ by (*), since
$$
[[[y_1] y_2] y_3]=[y_1 y_2 y_3]=e;
$$
the same $(*)$ also simplifies the right hand side of $(**)$ as
$$
y_1 [y_1]^{-1} \cdot [y_1] y_2 [y_1 y_2]^{-1} \cdot [y_1 y_2] y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3
$$ 
Now the induction step in general must be easy. 
A: If $G$ is finitely generated by $A\subset G$, equipping $G$ with the word metric, we can look at $G$ as a proper metric space $(G,d_{A})$. Then $H$ acts continuously (by isometries) on $(G,d_{A})$. Provoking Švarc–Milnor lemma, we get that $H$ is f.g and quasi-isometric to $G$ (looks like $G$ looking at both $G$ and $H$ from far distance).
A: Here's the topological argument.  The fact that $G$ is finitely generated means that $G=\pi_1(K)$ for $K$ a CW-complex with finite 1-skeleton.  Let $\widehat{K}$ be the covering space corresponding to $H$.  Then $H=\pi_1(\widehat{K})$, and $\widehat{K}$ also has finite 1-skeleton, so $H$ is finitely generated.
A: Hint: Suppose $G$ has generators $g_1, \ldots, g_n$. We can assume that the inverse of each generator is a generator. Now let $Ht_1, \ldots, Ht_m$ be all right cosets, with $t_1 = 1$. For all $i,j$, there is $h_{ij} \in H$ with $t_i g_j = h_{ij} t_{{k}_{ij}}$, for some $t_{{k}_{ij}}$. It's not hard to prove that $H$ is generated by all the $h_{ij}$.
